Question: Find the distance between the point ${(-2, 0)}$ and the line $\enspace {y = -\dfrac{1}{3}x + 6}\thinspace$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Solution: First, find the equation of the perpendicular line that passes through ${(-2, 0)}$ The slope of the blue line is ${-\dfrac{1}{3}}$ , and its negative reciprocal is ${3}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = 3x + b}\thinspace$ We can plug our point, ${(-2, 0)}$ , into this equation to solve for ${b}$ , the y-intercept. $0 = {3}(-2) + {b}$ $0 = -6 + {b}$ $0 + 6 = {b} = 6$ The equation of the perpendicular line is $\enspace {y = 3x + 6}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(0, 6)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(-2, 0)}$ and ${(0, 6)}$ gives us: $\sqrt{( {-2} - {0} )^2 + ( {0} - {6} )^2}$ $= \sqrt{( -2 )^2 + ( -6 )^2} = \sqrt{40} = 2\sqrt{10}$ The distance between the point ${(-2, 0)}$ and the line $\thinspace {y = -\dfrac{1}{3}x + 6}\enspace$ is $\thinspace2\sqrt{10}$.